Question

I'm trying to find the partial fraction decomposition of:

10x^2+2x-2/x^2(x+1)

I already found B=-2 and C=6

Plugging in the numbers, I get:

10x^2+2x-2=Ax(x+1)+(-2)(x+1)+(6)x^2

How do I solve this for A? Maybe this is just some algebra that I'm rusty on. But I'm having trouble with this. Help please?

Answers

so far, so good. You now have
10x^2+2x-2 = Ax(x+1)+(-2)(x+1)+(6)x^2
10x^2+2x-2 = Ax^2+Ax-2x-2+6x^2
10x^2+2x-2 = (A+6)x^2 - (A-2)x - 2
for those two polynomials to be identical, all the coefficients must match. That is

A+6 = 10
A-2 = 2
So, A=4
Luckily, those two equations give the same solution for A. I'm just curious how you found B and C without seeing how to get A. B I can see, since you wound up with

(A+C)x^2 + (A+B)x + B
To get B, I plugged in A=0. To get C I plugged in B=-1
No, I plugged in X=0 and X=-1

Related Questions